Transcript Document

Recap
In terms of relativistic momentum,
the relativistic total energy can be
expressed as followed
2
2
u
c p
E  m c ; p  m u  2  2
c
E
2
2
2 4
0
2
2
2
0
2
2
2 4 2

mc
m0 c E 
2
2 2 4
2 2

 E   m0 c 
 m0 c
2
2 2 

u
E c p 

1 2
c
2 4
0
2
2 4 Conservation of
energy-momentum1
0
E  p c m c
2
2 2
Reduction of relativistic kinetic
energy in the classical limit
The expression of the relativistic
kinetic energy
K  m0c  m0c
2
2
must reduce to that of classical one in the limit
u –> 0 when compared with c, i.e.
lim K relativistic 
u  c
p
2
classical
2m0
2
m0u
(
)
2
2
Expand  with binomial expansion
For u << c, we can always expand  in
terms of (u/c)2 as
 u
  1  2
 c
2



1/ 2
u2
u4
 1  2  termsof order 4 and higher
2c
c
K  mc  m0c  c ( 1)
2
2




m0u
1u
2
 m0 c 1 
 ...  1 
2
2
 
 2 c
2
2
2
i.e., the relativistic kinetic energy reduces
to classical expression in the u << c limit
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Example
An electron moves with speed u = 0.85c. Find
its total energy and kinetic energy in eV.
CERN’s picture: the circular accelerator
accelerates electron almost the speed of light
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Due to mass-energy equivalence, sometimes
we express the mass of an object in unit of
energy
Electron has rest mass m0 = 9.1 x 10-31kg
The rest mass of the electron can be
expressed as energy equivalent, via
m0 c2 = 9.1 x 10-31kg x (3 x 108m/s)2
= 8.19 x 10-14 J
= 8.19 x 10-14 x (1.6x10-19)-1 eV
= 511.88 x 103 eV = 0.511 MeV
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Solution
 First, find the Lorentz factor,  = 1.89
 The rest mass of electron, m0c2, is
0.5 MeV
 Hence the total energy is
E = mc2 =  (m0c2)= 1.89 x 0.5 MeV =
0.97 MeV
 Kinetic energy is the difference between
the total relativistic energy and the rest
mass, K = E - m0c2
= (0.97 – 0.51)MeV = 0.46 MeV6
Mass-energy equivalence:
potential energy
In some special case, a system has no potential energy nor kinetic
energy, e.g. two nucleons at rest and separated far apart
According to mass-energy equivalence
E = mc2 = K + E0
However, in general, U and K for a system are not zero, e.g. two
nucleons fused into one nucleus – potential energy will come into the
play in such a nucleus and cannot be ignored
In fact, not only does kinetic energy contribute to the relativistic
mass, m, to the system, but potential energy too
Hence, more generally,
E = mc2 = K + U + E0
Hence, generally the mass of a system, m, would have a contribution
not only from its kinetic energy but also from potential energy
Its relativistic mass m and the rest mass m0 will be different by an
amount
Dmc2 = (m - m0)c2 = K + U
(in previous lecture we have temporarily ignore the role of potential energy
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for the sake of simplicity)
Example: a compressed spring
Consider an uncompressed spring system with rest
mass m0 (We shall ignore the kinetic energy as we consider only rest
spring)
Uncompressed, Rest mass = m0, total relativistic energy
E = E0 =m0c2 (rest energy only)
Now, the spring is compressed by Dx by some external force
Compression caused by
external force, Fext
The work done by the external force will be converted into the
potential energy of the spring, according to conservation of
mechanical energy, Fext Dx = U. U will add to the total relativistic
energy of the spring system:
E = m0c2 --–> E = m0c2 + U 8
As the total energy, E, increases due to external U, the
relativistic mass of the spring will increase by some small
amount Dm due to mass-energy equivalence
the increase in the mass is simply Dmc2
=
U
E.g. if the spring constant k = 100N/m, and is compressed
by 10 cm, potential energy stored = U = kx2/2 = 0.5 J
This will contribute to the total relativistic mass of the spring
by a amount Dm = U/c2 = 5.56 x 10-18 kg – what a tiny
amount
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Binding energy
The nucleus of a deuterium comprises of
one neutron and one proton. Both
nucleons are bounded within the
deuterium nucleus
Neutron, mn
proton, mp
Nuclear fusion
Initially, the total
Energy = (mn+ mn)c2
U
After fusion, the total
energy = mdc2 + U
Deuterium, md
Analogous to exothermic process in
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chemistry
U is the energy that will be released when
a proton and a neutron is fused in a
nuclear reaction. The same amount of
energy is required if we want to separate
the proton from the neutron in a
deuterium nucleus
U is called the binding energy
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U can be explained in terms energy-mass
equivalence relation, as followed
For the following argument, we will ignore KE
for simplicity sake
Experimentally, we finds that mn + mp > md
By conservation of energy-momentum,
E(before) = E(after)
mnc2 + mpc2 + 0 = mdc2 + U
Hence, U = (mp + mn)c2 - mdc2 = Dmc2
The difference in mass between deuterium and
the sum of (mn + mn)c2 is converted into the
binding energy that binds the proton to the
neutron together
(opposite to the case of a compressed spring – in
deuterium its mass decreases)
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Example
mn= 1.008665u; mp= 1.007276u;
md= 2.013553u;
u = standard atomic unit = mass of 1/12 of the mass
of a 12C nucleus
= 1.66 x 10-27kg
= 1.66 x 10-27 x c2 J = 1.494 x 10-10 J
= 1.494 x 10-10/(1.6x10-19) eV
= 933.75 x 106 eV = 933.75 x MeV
•Hence the binding energy
U = Dmc2 = (mp + mn)c2 - mdc2
=0.002388u = 2.23 MeV
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SR finishes here…
We will go to the next topic …
Mainly on the quantum picture of light
and matter
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